Building on the Shallow Ice Approximation

Building on the Shallow Ice Approximation#

On previous page we derived expressions for the horizontal ice velocity as a function of depth $\zeta$ and for the depth-integrated flux. On this page we build on these results to derive expressions for the depth-averaged velocity and the vertical velocity as a function of $\zeta$.

Horizontal velocity#

To summarize the previous page, we showed that the horizontal velocity as a function of depth $\zeta$ is

\[ u(\zeta)= \frac{2A}{n+1} \left(\rho g \alpha \right)^n H^{n+1} \left( 1- \zeta^{n+1} \right), \]

where $A$ is the flow parameter in Glen’s flow law (the power-law, shear-thinning rheology assumed for ice), $n$ is the flow-law exponent, $\rho$ is the ice density, $g$ is the acceleration due to gravity, $\alpha$ is the surface slope, $H$ is the ice thickness, and $\zeta$ is the dimensionless depth coordinate defined as $\zeta = 1-z/H$.

We can get the horizontal velocity at the surface by evaluating this expression at $\zeta = 0$:

\[ u_s = u(0)= \frac{2A}{n+1} \left(\rho g \alpha \right)^n H^{n+1}, \]

which shows that

\[ u(\zeta) = u_s \left( 1- \zeta^{n+1} \right). \]

Mean velocity#

We can compute the mean horizontal velocity from this by integrating vertically and dividing by $H$:

\[ \overline{u} = \frac{2A}{n+2} \left(\rho g \alpha \right)^n H^{n+1}. \]

Note

On the previous page we integrated vertically to get the depth-integrated flux $q$, so you could also obtain the expression above by simply dividing our expression for $q$ by $H$.

This looks very similar the surface velocity. In fact, according the shallow ice approximation, the surface velocity and the depth-averaged velocity have a very simple relationship:

\[ \frac{u_s}{ \overline{u}} = \frac{n+2}{n+1}. \]

The surface velocity is always larger than the depth-averaged velocity, but by a smaller fraction for increasing nonlinear ice rheology. For example, if is linear, $n=1$ and $u_s/\overline{u} = 3/2$, or if $n=3$ and $u_s/\overline{u} = 5/4$. This can be understood mathematically by looking at the horizontal velocity shape function $1- \zeta^{n+1} $. The second term in the shape function controls how much the velocity deviates from the surface velocity. It deviates less for higher $n$ (just consider what the curves of $y=x^{n+1}$ look like between 0 and 1 for different values of $n$). This effect can be understood physically by remembering that we showed the that vertical shear stress increases linearly with depth:

\[ \tau_{zx} = -\rho g \alpha (z-H). \]

It is the rheology of the ice that determines how this linear increase in $\tau_{zx}$ with depth translates into shear strain. Therefore, it makes sense that in more non-linear ice, the deformation is concentrated where the shear stress is highest – the bottom – and the vertical gradient of $u$ is highest there, which is exactly what $1- \zeta^{n+1} $ does for higher $n$.

Vertical velocity#

Another extension to our expression for horizontal velocity from the previous page,

\[ u(x, \zeta)= u_s(x) \left( 1- \zeta^{n+1} \right), \]

is to compute the vertical velocity as function of depth. This is often referred to as the Lliboutry function. In the expression above we are explicitly noting that $u_s$ and therefore $u$ are functions of $x$. We will use the general mass continuity equation,

$ \nabla\cdot\underline{u} = \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0, $$

where $w$ is the vertical velocity, $u$ is the horizontal velocity, $x$ is the horizontal coordinate, and $z$ is the vertical coordinate. As our velocity expression is in terms of normalized depth, $\zeta = 1-z/H$, we want the expression above in terms of $\zeta$. Differentiating the definition of $\zeta$ provides

\[ \frac{\partial \zeta}{\partial z} = -\frac{1}{H} \]

and applying the chain rule gives

\[ \frac{\partial w}{\partial z} = \frac{\partial w}{\partial \zeta} \frac{\partial \zeta}{\partial z} = -\frac{1}{H} \frac{\partial w}{\partial \zeta}. \]

Therefore, from the mass continuity equation

\[ \frac{\partial u}{\partial x} - \frac{1}{H} \frac{\partial w}{\partial \zeta} = 0. \]

Differentiating our expression for $u(x, \zeta)$ with respect to $x$ gives

\[ \frac{\partial u}{\partial x} = \frac{\partial u_s}{\partial x} \left( 1- \zeta^{n+1} \right) - u_s \frac{\partial \zeta}{\partial x} (n+1) \zeta^n. \]

Here we assume that $H$ is not a function of $x$, so $\partial \zeta/\partial x = 0$. Putting this into the expression above leaves

\[ \frac{\partial u}{\partial x} = \frac{\partial u_s}{\partial x} \left( 1- \zeta^{n+1} \right). \]

Based on the mass continuity equation

\[ \frac{\partial u_s}{\partial x} \left( 1- \zeta^{n+1} \right) = \frac{1}{H} \frac{\partial w}{\partial \zeta}. \]

Rearranging this gives

\[ \frac{\partial w}{\partial \zeta} = H \frac{\partial u_s}{\partial x} \left( 1- \zeta^{n+1} \right). \]

Now we need to integrate vertically to get $w(\zeta)$. First we put in the limits of integration to define the boundary conditions. At the base, $\zeta = 1$ and $w=w_b$.

\[ \int^w_{w_b} dw = H \frac{\partial u_s}{\partial x} \int^\zeta_1\left( 1- \zeta^{n+1} \right) \]

Evaluating the integrals gives

\[ w - w_b = H \frac{\partial u_s}{\partial x} \left[\zeta - \frac{\zeta^{n+2}}{n+2} \right]^\zeta_1 = H \frac{\partial u_s}{\partial x} \left[\zeta - \frac{\zeta^{n+2}}{n+2} - 1 + \frac{1}{n+2} \right]. \]

Noting that

\[ - 1 + \frac{1}{n+2} = - \frac{n+2}{n+2} + \frac{1}{n+2} = \frac{-n-2+1}{n+2} = -\frac{n+1}{n+2} \]

gives

\[ w - w_b = H \frac{\partial u_s}{\partial x} \left[\zeta - \frac{\zeta^{n+2}}{n+2} -\frac{n+1}{n+2} \right]. \]

We can define the vertical velocity at the surface ($\zeta=0$) as

\[ w_s - w_b = - H \frac{\partial u_s}{\partial x} \frac{n+1}{n+2}. \]

we can divide our expression for simplify our expression $w - w_b$ by dividing to

\[ w - w_b = (w_s - w_b) \left(1 -\frac{n+2}{n+1}\zeta + \frac{1}{n+1}\zeta^{n+2} \right). \]

These are the expressions for the vertical velocity as a function of depth $\zeta$ and the vertical velocity at the surface.

Perhaps easier to understand is the case when $w_b = 0$, i.e. the ice is not melting or being accreted at the base:

\[ w_s = - H \frac{\partial u_s}{\partial x} \frac{n+1}{n+2}, \]

\[ w = w_s \left(1 -\frac{n+2}{n+1}\zeta + \frac{1}{n+1}\zeta^{n+2} \right). \]

Consider the first of the above two expressions. The thickness$H$ and the flow exponent $n$ are positive, so the vertical velocity has the opposite sign to horizontal strain rate at the surface. This makes sense because the horizontal extension should result in vertical compression. In a place where $w_b=0$ this corresponds to a negative (i.e. downward) vertical velocity: think of a spring, initially stretched vertically, is slowly released - if the bottom end is held static (i.e. $w_b=0$) the whole spring is slowly moving downwards, albeit at spatially varying rates.

Now consider the seccond expression above. At the surface $\zeta=0$ and $w=w_s$ as required. The expression in the brackets is a so called shape function. The value of the flow exponent controls its shape. In the case when $n$ is very large, approximating a plastic rheology,

\[ w = w_s \left(1 -\zeta \right). \]

In this case, the vertical velocity varies uniformly from $w_s$ at the surface to zero at the base. This is because all the vertical shear is concentrated in a very small layer at the bed and the horizontal strain rate $\frac{\partial u_s}{\partial x}$ is uniform throughout the thickness of the ice.

As $n$ decreases and the ice behaves less plastically, and more viscously, the vertical shear spreads out through the thickness of the ice and the vertical velocity profile become nonlinear.